I have to say if following statements are true or false. For the initial value problem: $x'=f(x), x(t_0)=x_0$ with $f:\mathbb{R^n}\rightarrow\mathbb{R^n}$ , $t_0\in\mathbb{R},x_0\in\mathbb{R^n}$

  1. If $f$ is continuous and bounded then for all $t_0\in\mathbb{R},x_0\in\mathbb{R^n}$ there exists a global solution.
  1. If $f$ is a linear function then for all $t_0\in\mathbb{R},x_0\in\mathbb{R^n}$ there exists a unique global solution.

I have a small problem with the concept of a global solution. I know from the Peano Theorem that if $f$ is continuous then there exists a local solution, if $f$ is locally Lipschitz continuous, then this solution is unique(not sure if locally or everywhere). Then I think that if we know that it is locally Lipschitz and continuous we can expand the solution to the maximum existence interval. And if this interval is the whole domain then we can say the solution is global. Is this correct? What are other things I have to know so that I can say if it has a global solution or not and especially if this solution is unique?

Thanks a lot in advance


1 Answer 1


Take your insight into the maximal solution, with the given domain of the full time axis times the full state space, gives the statement that no solution can end at a point that has all coordinates finite, time and space.

Now if the function $f$ is linearly bounded, one finds that the solutions are exponentially bounded, so a divergence to space infinity is not possible in finite time. It remains that the time coordinate extends to infinity.

  • $\begingroup$ Thanks for the answer. So the statement is true, for both of them? cause they are two different statements and would be great if you could expand a bit more on your answer. $\endgroup$
    – Annalisa
    Commented Feb 20, 2023 at 19:05
  • $\begingroup$ Yes. In the first case you get a linear bound, in the second an exponential bound for the solution of the IVP. Both prevent divergence in finite time. $\endgroup$ Commented Feb 21, 2023 at 8:59

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